Let \(R\) be a commutative ring and let \(M\) and \(N\) be \(R\)-modules. Prove the following.
If \(M\) and \(N\) are projective \(R\)-modules, then \(M \otimes_R N\) is a projective \(R\)-module.
If \(M\) and \(N\) are flat \(R\)-modules, then \(M \otimes_R N\) is a flat \(R\)-module.
If \(M\) is a flat \(R\)-module and \(N\) is an injective \(R\)-module, then \(\operatorname{Hom}_R(M, N)\) is an injective \(R\)-module.
Let \(p\) be a prime number and let \(\mathbb{Z}_{(p)}\) denote the localization of \(\mathbb{Z}\) with respect to its prime ideal \(p\mathbb{Z}\). Show that \(\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}_{(p)}, \mathbb{Q}/\mathbb{Z})\) is an injective \(\mathbb{Z}\)-module.
Solution
1. If \(F\) and \(G\) are free \(R\)-modules, then \(F\otimes_R G\) is free.
Write \[ F\cong\bigoplus_{i\in I}R, \qquad G\cong\bigoplus_{j\in J}R. \] Tensor product commutes with direct sums in each variable, so \[ F\otimes_R G \cong \bigoplus_{(i,j)\in I\times J}(R\otimes_R R) \cong \bigoplus_{I\times J}R, \] which is free.
2. If \(M\) and \(N\) are projective, then \(M\otimes_RN\) is projective.
Choose modules \(M'\) and \(N'\) and free modules \(F,G\) such that \[ F\cong M\oplus M', \qquad G\cong N\oplus N'. \] Then \[ F\otimes_R G \cong (M\otimes_RN) \oplus(M\otimes_RN') \oplus(M'\otimes_RN) \oplus(M'\otimes_RN'). \] By 1, \(F\otimes_R G\) is free. Hence \(M\otimes_RN\) is a direct summand of a free module and therefore projective. This proves part (a).
3. If \(M\) and \(N\) are flat, then \(M\otimes_RN\) is flat.
For every \(R\)-module \(X\), associativity of tensor product gives a natural isomorphism \[ X\otimes_R(M\otimes_RN) \cong (X\otimes_RM)\otimes_RN. \] Thus the functor \[ -\otimes_R(M\otimes_RN) \] is naturally isomorphic to the composite \[ (-\otimes_RM)\quad\text{followed by}\quad(-\otimes_RN). \] Both functors are exact because \(M\) and \(N\) are flat, so their composite is exact. Hence \(M\otimes_RN\) is flat. This proves part (b).
4. Let \(M\) be flat and \(N\) injective. Then \(\operatorname{Hom}_R(M,N)\) is injective.
Let \(A\hookrightarrow B\) be an injective homomorphism of \(R\)-modules. Flatness of \(M\) makes \[ A\otimes_RM\longrightarrow B\otimes_RM \] injective. Since \(N\) is injective, every homomorphism \[ A\otimes_RM\longrightarrow N \] extends to \(B\otimes_RM\).
By the tensor–Hom adjunction, \[ \operatorname{Hom}_R(A,\operatorname{Hom}_R(M,N)) \cong \operatorname{Hom}_R(A\otimes_RM,N), \] and similarly with \(B\) in place of \(A\). Therefore every map \[ A\longrightarrow\operatorname{Hom}_R(M,N) \] extends to \(B\). This is precisely the injectivity of \(\operatorname{Hom}_R(M,N)\). This proves part (c).
5. The \(\mathbb Z\)-module \(\mathbb Z_{(p)}\) is flat.
Every localization \(S^{-1}R\) is flat as an \(R\)-module. Here \[ \mathbb Z_{(p)}=S^{-1}\mathbb Z, \qquad S=\mathbb Z\setminus p\mathbb Z. \]
6. The \(\mathbb Z\)-module \(\mathbb Q/\mathbb Z\) is injective.
A \(\mathbb Z\)-module is injective if and only if it is divisible. The group \(\mathbb Q/\mathbb Z\) is divisible: given \(q+\mathbb Z\) and a nonzero integer \(n\), \[ n\left(\frac qn+\mathbb Z\right)=q+\mathbb Z. \] Hence \(\mathbb Q/\mathbb Z\) is injective.
7. Therefore\[ \operatorname{Hom}_\mathbb Z(\mathbb Z_{(p)},\mathbb Q/\mathbb Z) \]is an injective \(\mathbb Z\)-module.
6. This proves part (d).